If the sin Alfa and cos Alfa are root of equation a^2+bx+c then prove

1.	If the sin Alfa and cos Alfa are root of equation a^2+bx+c then prove that a^2+2ac=b^2
Answer» The given equation is {tex}ax^2 + bx + c = 0{/tex}sin\xa0{tex}\\alpha{/tex}\xa0and cos\xa0{tex}\\alpha{/tex}\xa0are roots of the given equation.{tex}\\therefore{/tex}\xa0sin\xa0{tex}\\alpha{/tex}\xa0+ cos\xa0{tex}\\alpha{/tex}\xa0=\xa0-{tex}\\frac{b}{a}{/tex}\xa0...(i)and {tex}sin{/tex}\xa0{tex}\\alpha{/tex}{tex}.cos{/tex}\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{c}{a}{/tex}...(ii)Squaring both sides of equation (i),\xa0{tex}\\Rightarrow {/tex}(sin\xa0{tex}\\alpha{/tex}\xa0+ cos\xa0{tex}\\alpha{/tex})2 =\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}sin^2{/tex}\xa0{tex}\\alpha{/tex}\xa0{tex}+ cos^2{/tex}\xa0{tex}\\alpha{/tex}\xa0+\xa02 sin\xa0{tex}\\alpha{/tex}cos\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}1 + 2{tex}\\frac{c}{a}{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{a + 2c}{a}{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}a^2 + 2ac = b^2{/tex}

If the sin Alfa and cos Alfa are root of equation a^2+bx+c then prove that a^2+2ac=b^2